(i) Values displayed are: Creative mind, awareness, scientific interest and keen observer.
(ii) Here v = 350 m/s, l = 0.5 m
∴ For closed pipe fundamental frequency
\(v=\frac{v}{4l}=\frac{350}{4\times 0.5}=175 Hz\)
Frequency of second overtone = 5v=\(5 \times 175 = 875 Hz\)

Fundamental frequency of open pipe is
\(v_{0} = \frac{v}{2l}\)  --- (i)
Frequency of second harmonic of closed pipe of same length is
\(v_{c}=\frac{3v}{4l}=\frac{3}{2}(\frac{v}{2l})=\frac{3}{2} v_{0}\)
Given v_c=v₀+100
⇒ \(\frac{3}{2} v_0 -v_0=100\)
or  \(\frac{v_0}{2}=100\)
∴ v₀= 200 Hz.

(a) v^'=2v.
It is possible is source is approaching the stationary listener i.e., v_s is +
As \(v^{'}=\frac{v-v_{L}}{v-v_{s}}V\)
∴ 2v =  \(\frac{v}{v-v_{s}}V\)
or 2 = \(\frac{v}{v-v_{s}}\)
or 2v - 2v_s = v
or 2v_s = v
or \(v_{s}=\frac{v}{2}\)
Therefore, source should approach the listener with half the velocity of sound propagation.
(b) \(v^{'}=\frac{v}{2}\)
It is possible if source is receding away from the stationary listener i.e., v_s is negative and v_L= 0
∴ \(v^{'}=\frac{v}{v+v_{s}}v\)
or \(\frac{v}{2}=\frac{v}{v+v_{s}}v \ or\ \frac{1}{2}=\frac{v}{v+v_{s}}\)
or v + v_s = 2v or v_s = v
Therefore, source should recede away from the listener with the velocity of sound propagation.

Let us consider the case of three disturbances each of amplitude a and frequencies (n -1), and (n + 1) respectively. The resultant displacement is given by
y = a sin 2ㅠ(n-1)t + a sin 2ㅠnt+a sin 2ㅠ(n+1)t
= 2a sin 2ㅠnt cos2ㅠnt + a sin 2ㅠnt
= a(1 + 2cos 2ㅠt)sin 2ㅠnt 
So the resultant amplitude is a (1 + 2cos 2ㅠt)
which is maximum when cos 2ㅠt = + 1
ஃ  2ㅠt = 2k where k = 0, 1, 2, 3 .........
t = 0, 1, 2, 3 ...
Thus the time interval between two consecutive maxima is one. This shows that the frequency of maxima is one.
Similarly, the amplitude is minimum when
1 + 2 cos 2 π t = 0
or \(cos 2\pi t= -\frac{1}{2}\)
or \(2 \pi t= 2k \pi+\frac{2\pi}{3} \)   (where k=0,1,2,...)
or \(t=(k+\frac{1}{3})\)
=\(\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3}\)
Thus the minima occurs after an interval of one second, i.e., the frequency of minima is also one. Hence, the frequency of beats is also one.
Thus, one beat is heard per second.

Suppose velocity of sound in air at t⁰C is 1\(\frac{1}{2}\) times the velocity at 11°C.
i.e., \(v_{t}=\frac{3}{2} v_{11}\)
As \(v_{t}=v_{0} \sqrt{\frac{273+t}{273}}\)
ஃ from (i), \(v_{0}=\sqrt{\frac{273+t}{273}}=\frac{3}{2}v_{11}\)
\(v_{0}=\sqrt{\frac{273+t}{273}}=\frac{3}{2}\ v_{0}\frac{284}{273}\)
Squaring both sides, we get
\(\sqrt{\frac{273+t}{273}}=\frac{9}{4} \times \frac{284}{273}\)
or 1092 + 4t = 2556
or 4t = 2556 - 1092 = 1464
or t = \(\frac{1464}{4}= 360 ^{0}C\)

\(v_{1}=\sqrt{\frac{\gamma_{1}P}{\rho_{1}}}\ and \ J_{2}=\sqrt{\frac{\gamma_{2}P}{\rho_{2}}}\)
\(\frac{v_{1}}{v_{2}}=\sqrt{\frac{\gamma_{1}\rho_{2}}{\gamma_{2} \rho_{1}}}\)
Since density of a gas is proportional to its molecular weight,
\(\frac{\rho_{2}}{\rho_{1}}=\frac{44.01}{2.016}=21.83\)
\(\frac{v_{1}}{v_{2}}=\sqrt{\frac{1.4}{1.3}\times 21.83}=4.85\)
Velocity of sound in hydrogen in 4.85 times that in carbon dioxide.

The given function is y(x,t) = A sin(kt - ωt)+  B cos (kx - ωt).
Let us put \(A= a cos \phi \ and \ B=a sin \phi\), where \(a=\sqrt{A^{2}+B^{2}}\) and \(tan \phi=\frac{B}{A}\).Then, the above function may be expressed as 
\(y(x,t)=a cos \phi sin (kx-\phi t)+a sin \phi cos(kx-\phi t)\)
\(y(x,t)= a sin (kx-\omega t+\phi)\)
As it is a single sinusoidal function of space and time, it is representing the equation of a harmonic progressive wave. 
Amplitude of wave = \(a=\sqrt{A^{2}+B^{2}}\)
Wavelength of wave= \(\lambda = \frac{2\pi}{k}\).
Wave velocity = v = vλ = \(\frac{v}{2\pi}.\frac{2\pi}{k}=\frac{v}{k}\).
and initial phase angle of the wave = \(\phi\) where, \(\phi tan^{-1}(\frac{B}{A})\).

As   2 sin A cos B sin (A + B) + sin (A - B)
y = 2A sin kx cos ωt
= A sin(kx + ωt) + A sin(kx - ωt)
According to superposition principle,
y = y₁ + y₂;
and y₁ =  A sin (ωt-kx)
= - A sin(kx - ωt)
∴ y₂ = y - y₁ = 2A sin kx cos ωt + A sin(kx-ωt)
= A sin(kx + ωt) + 2A sin(kx - ωt)
= A sin(kx + ωt) - 2A sin(ωt - kx).

The basic conditions for formation of stationary waves are listed below:
(i) The direct and reflected waves must be travelling along the same line.
(ii) For stationary wave formation, the superposing waves should either be longitudinal or transverse. A longitudinal and a transverse wave cannot superposition.
(iii) For formation of stationary waves, there should not be any relative motion between the medium and oppositely travelling waves.
(iv) Amplitude and period of the superposing waves should be same.

According to the given problem the frequency of the last fork is the octave of the first, i.e., if the frequency of first fork is n, then the frequency of last fork is 2n. This shows that the forks are in increasing frequency order. As each fork gives 3 beats with previous are, hence frequencies are
n, (n+3), (n+2\(\times\)3), (n+3\(\times\)3), .... 2n
In forms an A.P.
∴ an = a + (n-1)d
or 2n = n + (65-1)\(\times\)3 
By solving, we get n = 192 and 2n = 384
Frequency of first and last forks are 192 and 384 respectively.

v = 256 Hz, T= m x g, T = 90 x 9.8 = 882N
\(\mu =\frac { m }{ L } =8.0\times 10^{ -3 }kgm^{ -1 }\)
Amplitude, a = 5 cm = 0.05 cm
Velocity of the transverese wave
\(\Rightarrow v=\sqrt { \frac { T }{ \mu } } =\sqrt { \frac { 882 }{ 8\times 10^{ -3 } } } =3.32\times 10^{ 2 }m/s\)
\(\\ \omega =2\pi v=2\times 3.14\times 256\)
\(=\ 1.61\times 10^{ 3 }rad/s\)
\(\lambda =\frac { v }{ V } =\frac { 3.32\times 10^{ 2 } }{ 256 }\)
\( \\ k=\frac { 2\pi }{ \lambda } =\frac { 2\times 3.14\times 256 }{ 3.32\times 10^{ 2 } } =4.84m^{ -1 }\)
As wave propagation along positive X-axis
\(Y=\ asin(\omega t-kx)\)
\( =\ 0.05sin(1.61\times 10^{ 3 }t-4.84x)\)
Here x,y are in metre and t is in second.

The speed of sound wave in each case will be same and is 340 m/s.

\(\frac { \rho { A }^{ 2 }{ \varpi }^{ 2 }\pi s }{ k } \)

The fundamental frequency of a string fixed at both ends is given by
\(v=\frac{1}{2 L} \sqrt{\frac{F}{\mu}}\)
As F and mu are fixed, \(\frac{v_1}{v_2}=\frac{L_2}{L_1}$ or , $L_2=\frac{v_1}{v_2} L_1=\frac{124 \mathrm{~Hz}}{186 \mathrm{~Hz}}(90 \mathrm{~cm})=60 \mathrm{~cm}\).
Thus, the string should be pressed at 60 cm from an end.